Problem: The equation of hyperbola $H$ is $\dfrac{y^2}{16}-\dfrac {(x-6)^{2}}{36} = 1$. What are the asymptotes?
Solution: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac{y^2}{16} = 1 + \dfrac {(x-6)^{2}}{36}$ Multiply both sides of the equation by $16$ $y^2 = { 16 + \dfrac{ (x-6)^{2} \cdot 16 }{36}}$ Take the square root of both sides. $\sqrt{y^2} = \pm \sqrt { 16 + \dfrac{ (x-6)^{2} \cdot 16 }{36}}$ $ y = \pm \sqrt { 16 + \dfrac{ (x-6)^{2} \cdot 16 }{36}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y \approx \pm \sqrt {\dfrac{ (x-6)^{2} \cdot 16 }{36}}$ $y \approx \pm \left(\dfrac{4 \cdot (x - 6)}{6}\right)$ Rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{2}{3}(x - 6)+ 0$